Q61 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

If $(-c, c)$ is the set of all values of $x$ for which the expansion of $(7-5 x)^{-2 / 3}$ is valid, then $5 c+7=$

  • A
    $0$
  • B
    $12$
  • C
    $41$
  • Correct answer
    D
    $14$

Solution

$(7-5 x)^{\frac{-2}{3}}$ $$ =7^{\frac{-2}{3}}\left(1-\frac{5 x}{7}\right)^{-\frac{2}{3}} $$ To make expansion valid $$ \begin{aligned} & \left|\frac{5 x}{7}\right| < 1 \Rightarrow|x| < \frac{7}{5} \\ & \therefore x \in\left(\frac{-7}{5}, \frac{7}{5}\right) \\ & \therefore C=\frac{7}{5} \therefore 5 C+7=14 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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